uva 10104 （扩展欧几里德算法）

    今天粗略学习了扩展欧几里德算法。

  设a,b,c为任意整数，g=gcd(a,b),方程 ax+by=g的一组解是（x1,y1）,则当c是g的倍数时，ax+by=c的一组解是(x1*c/g,y1*c/g);当c不是g的倍数时无整数解。

UVA 10104：

he Problem

From Euclid it is known that for any positive integers A and B there exist such integers X and Y thatAX+BY=D, where D is the greatest common divisor of A and B. The problem is to find for given A and Bcorresponding X, Y and D.

The Input

The input will consist of a set of lines with the integer numbers A and B, separated with space (A,B<1000000001).

The Output

For each input line the output line should consist of three integers X, Y and D, separated with space. If there are several such X and Y, you should output that pair for which |X|+|Y| is the minimal (primarily) and X<=Y (secondarily).

Sample Input

4 617 17

Sample Output

-1 1 20 1 17

代码：

#include<cstdio>
void gcd(int a,int b,int &d,int &x,int &y)
{
    if(b==0)
    {
        d=a;
        x=1;y=0;
    }
    else
    {
        gcd(b,a%b,d,y,x);
        y-=x*(a/b);
    }
}
using namespace std;
int main()
{


    int a,b,d,x,y;
    while(scanf("%d%d",&a,&b)!=EOF)
    {
         gcd(a,b,d,x,y);
         printf("%d %d %d\n",x,y,d);
    }
    return 0;
}